A356783 Coefficients in the power series A(x) such that: 1 = Sum_{n=-oo..+oo} x^(2*n+1) * (1 - x^n)^(n+1) * A(x)^n.
1, 1, 2, 6, 17, 50, 163, 525, 1770, 6066, 21154, 74787, 267371, 965233, 3513029, 12877687, 47499333, 176167086, 656568385, 2457710598, 9236079055, 34832753818, 131792634266, 500121476517, 1902979982421, 7258942377746, 27752992782498, 106333425162358, 408213503595652
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 17*x^4 + 50*x^5 + 163*x^6 + 525*x^7 + 1770*x^8 + 6066*x^9 + 21154*x^10 + 74787*x^11 + 267371*x^12 + ... such that 1 = ... + x^(-3)*(1 - x^(-2))^(-1)/A(x)^2 + x^(-1)/A(x) + x*0 + x^3*(1 - x)^2*A(x) + x^5*(1 - x^2)^3*A(x)^2 + x^7*(1 - x^3)^4*A(x)^3 + ... + x^(2*n+1)*(1 - x^n)^(n+1)*A(x)^n + ... also -A(x)^3 = ... + x^(-3)*(A(x) - x^(-2))^(-1)*A(x)^2 + x^(-1)*A(x) + x*(A(x) - 1) + x^3*(A(x) - x)^2/A(x) + x^5*(1 - x^2)^3/A(x)^2 + x^7*(A(x) - x^3)^4/A(x)^3 + ... + x^(2*n+1)*(A(x) - x^n)^(n+1)/A(x)^n + ...
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..400
Crossrefs
Programs
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PARI
{a(n) = my(A=[1]); for(i=0,n, A = concat(A,0); A[#A] = polcoeff(1 - sum(n=-#A\2-1,#A\2+1, x^(2*n+1) * (1 - x^n +x*O(x^#A))^(n+1) * Ser(A)^n ),#A-2); );A[n+1]} for(n=0,30, print1(a(n),", "))
Formula
G.f. A(x) = Sum_{n>=0} a(n) * x^n satisfies the following relations.
(1) 1 = Sum_{n=-oo..+oo} x^(2*n+1) * (1 - x^n)^(n+1) * A(x)^n.
(2) x*A(x) = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n-1)) / ( (1 - x^(n+1))^n * A(x)^n ).
(3) -x*A(x)^2 = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n-1)) * A(x)^n / (1 - x^(n+1)*A(x))^n.
(4) -A(x)^3 = Sum_{n=-oo..+oo} x^(2*n+1) * (A(x) - x^n)^(n+1) / A(x)^n.
(5) 0 = Sum_{n=-oo..+oo} x^(2*n+1) * (1 - x^n*A(x))^(n+1) / A(x)^n.
(6) 0 = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n-1)) * A(x)^n / (A(x) - x^(n+1))^n.
a(n) ~ c * d^n / n^(3/2), where d = 4.04962821886295599791727073173857... and c = 0.613483546803830745310382482744... - Vaclav Kotesovec, Mar 22 2025
Comments